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FRQ: Vertical Circular Motion — Inside the Loop

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points

▤ Scenario

A stunt rider and motorcycle (combined mass 200 kg) travel around the inside of a vertical circular loop of radius 6.4 m at a stunt show. The rider enters the loop at the bottom and completes it. The speed of the motorcycle is not constant around the loop. Take g = 10 m/s². Rolling friction and air resistance are negligible, so no energy is lost to them.

Motion occurs in a vertical plane. Use an inertial reference frame fixed to the ground. On the radial axis at any point of the loop, take the direction toward the centre of the loop as positive.

✎ Free Response Questions

(a) Draw (or describe completely) the free-body diagram of the rider–motorcycle system (i) at the top of the loop and (ii) at the bottom of the loop. Name each force, name the object exerting it, and give its direction. For the bottom of the loop, also indicate which of the two forces is larger.

(b) Starting from Newton's second law applied at the top of the loop, derive a symbolic expression for the minimum speed v_min at the top required to maintain contact with the track. Then calculate v_min for this loop.

(c) At the bottom of the loop the motorcycle moves at 20 m/s. Calculate the magnitude of the normal force exerted by the track on the rider–motorcycle system at that instant, and compare it to the system's weight (express the comparison as a ratio).

(d) Rider A (total mass 200 kg) and Rider B, who rides a lighter machine (total mass 150 kg), argue before the stunt. Rider A claims: “I am heavier, so I need a higher speed at the top to avoid falling.” Rider B claims that both riders need the same minimum speed. In a paragraph-length response, state which rider is correct and construct a coherent argument that connects your symbolic result from part (b) and the numerical value you calculated there to the physical reason the two riders do or do not differ. Your argument must reference both the force available to curve the path and the net force required for circular motion.

❖ Answer Key & Scoring Guide

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part (a) — Model Answer

(i) At the top of the loop the system is in contact with the track and exactly two forces act on it, both directed vertically downward — which at that location is toward the centre of the loop: the gravitational force F_g = mg exerted by the Earth, and the normal force F_N exerted by the track. The track lies above the system there, and a normal force can only push perpendicular to and away from the surface — it can never pull.

(ii) At the bottom of the loop the gravitational force F_g = mg exerted by the Earth points downward, away from the centre, and the normal force F_N exerted by the track points upward, toward the centre. Since the net force must point toward the centre at that location, F_N is necessarily larger than mg.

At both locations the net force points toward the centre of the loop. No separate “centripetal force” is exerted on the system by anything; that name belongs to the net of the real forces already drawn.

Free-body diagrams of the rider–motorcycle system at the top and the bottom of the loop. The dashed grey arc marks the track surface and the dashed amber arrow marks the positive radial direction — toward the centre — at each location. In the top diagram the arrow lengths show direction only: which of F_N and F_g is larger there depends on the speed at the top.

Scoring (2 points):

▸ 1 point: Top-of-loop diagram showing the gravitational force and the normal force both directed downward, with no upward normal force, and each force named together with the object exerting it (the Earth; the track).

▸ 1 point: Bottom-of-loop diagram showing the gravitational force downward and the normal force upward, each named together with the object exerting it, and F_N drawn or stated to be larger than mg.

⚠︎ Common error (partial credit): drawing the normal force upward at the top of the loop — earns 0 of 1 for the top diagram, but the bottom diagram is scored independently and its point remains available. A surface can push away from itself; it can never pull.

⚠︎ Common error (partial credit): a top diagram showing only F_g, explicitly justified as the limiting case in which the system travels at its minimum speed and F_N = 0, earns the first point; the same single-arrow diagram with no such justification earns 0 of 1 for that diagram.

✗ Common error (no credit): adding a third arrow labelled “centripetal force” to a diagram — that diagram earns 0 of its 1 point, and the other diagram is still scored independently. The centripetal force is the name for the net of the real forces already drawn, not an additional force exerted by any object.

Part (b) — Model Answer

At the top of the loop, taking toward the centre — downward at that location — as positive, Newton's second law along the radial axis gives:

F_N + mg = m v² / r

The system stays in contact with the track only while F_N ≥ 0. The minimum speed is the limiting case F_N = 0, in which the gravitational force alone provides the net force required for circular motion:

mg = m v_min² / r  ⇒  v_min = √(g r)

v_min = √(10 m/s² × 6.4 m) = √(64 m²/s²) = 8.0 m/s

The minimum speed at the top of the loop is 8.0 m/s. Below this speed the net force required for circular motion, m v² / r, is smaller than the gravitational force mg. Restoring the balance would require the track to pull the system outward — upward at that location — and a surface can only push. The inward force is then larger than a circle of radius r demands, so the system curves inside the track and loses contact with it.

Scoring (2 points):

▸ 1 point: Begins from Newton's second law rather than from a rearranged result — writes F_N + mg = m v² / r at the top, with both forces taken toward the centre. This point is awarded for the correct starting equation alone, before any algebra is performed.

▸ 1 point: Applies the contact condition F_N = 0 (or F_N ≥ 0) to obtain the symbolic result v_min = √(g r), and evaluates it correctly as 8.0 m/s with units.

⚠︎ Common error (partial credit): setting F_N = mg at the top, transferring the flat-ground result, gives v = √(2 g r) ≈ 11.3 m/s — earns the first point only, and only if the second-law equation was written correctly before the wrong condition was imposed.

⚠︎ Common error (partial credit): writing mg = m v_min² / r directly, with no mention of F_N and no statement of the contact condition — earns the second point, since the symbolic result and the value are correct, but not the first, because the part asks the student to start from Newton's second law.

Part (c) — Model Answer

At the bottom of the loop, taking toward the centre — upward at that location — as positive, Newton's second law along the radial axis gives:

F_N − mg = m v² / r  ⇒  F_N = m (g + v² / r)

F_N = 200 kg × (10 m/s² + (20 m/s)² / 6.4 m) = 200 kg × 72.5 m/s² = 14500 N

The normal force is 14500 N, directed upward. The weight of the system is mg = 2000 N, so F_N / mg = 7.25: the track pushes on the system with 7.25 times the system's weight. That is why a rider feels heaviest at the bottom of a loop — and why real stunt loops are built with a larger radius, since a 6.4 m loop taken at this speed loads the rider close to what a person can briefly tolerate.

Scoring (2 points):

▸ 1 point: Correct second-law equation at the bottom, with F_N upward, mg downward and the net force toward the centre: F_N − mg = m v² / r. This point is awarded for the correct starting equation alone.

▸ 1 point: Correct magnitude F_N = 14500 N with units and the correct comparison F_N / mg = 7.25, or the equivalent statement “7.25 times the system's weight”. A value and a ratio that both follow consistently from the student's own radial equation also earn this point.

⚠︎ Common error (partial credit): writing F_N = m v² / r alone, omitting gravity, gives 12500 N and a ratio of 6.25 — the equation point is lost, but the second point is earned if both the value and the ratio follow consistently from that incorrect equation.

✗ Common error (no credit): reporting F_N = mg = 2000 N because the surface “holds the system up”. At the bottom of the loop the system is accelerating toward the centre, so the forces are not balanced and F_N must exceed mg.

Part (d) — Model Answer

Rider B is correct: both riders need the same minimum speed at the top of the loop, 8.0 m/s. The symbolic result from part (b), v_min = √(g r), contains only g and r — the mass has already cancelled before the numerical stage, which is why substituting 200 kg or 150 kg makes no difference to the 8.0 m/s value. The physical reason is that mass enters the contact condition twice, once on each side of the same equation. The force available to curve the path at the top, in the limiting case, is the gravitational force mg, which is proportional to mass. The net force required to hold the system on a circle of radius r at speed v is m v² / r, which is proportional to mass in exactly the same way. Going from Rider A's 200 kg to Rider B's 150 kg reduces the gravitational force available by a quarter, but it reduces the net force required for the same circular path by exactly the same quarter, so the two changes cancel and the limiting speed is unchanged. Rider A's intuition holds the required force fixed while letting only the available force grow with mass; in fact both scale together. This is the same cancellation that makes free-fall acceleration independent of mass.

Scoring (2 points):

▸ 1 point: Selects Rider B and supports the claim with the symbolic result v_min = √(g r), noting explicitly that the mass does not appear in it. A student who carried an incorrect v_min forward from part (b) earns this point by applying that value consistently to both riders.

▸ 1 point: Paragraph-length physical reasoning naming both scalings — the available gravitational force mg and the required net force m v² / r are each proportional to mass — and concluding that mass therefore cancels in the contact condition.

⚠︎ Common error (partial credit): “Rider B is correct because the formula has no m in it”, with no physical reasoning — earns the first point only. Claim and evidence are present, but the reasoning that links the two mass-proportional forces is absent.

✗ Common error (no credit): selecting Rider A because a heavier system “needs more force” to travel the same circle. That statement is true but irrelevant on its own: the gravitational force supplying the requirement grows by exactly the same factor.